Hello, how can I make use of mpmath.besselj(n,x) with w a symbol?
as in result = r[a]*mpmath.sqrt(mpmath.pi/2)*(Symbol("C0")
*mpmath.besselj(0.5,x) + Symbol("C1")*mpmath.bessely(0.5,x))/
mpmath.sqrt(x)
where x = Symbol("x")
I get the follwoing error:
/home/cohen/data1/sources/python/sympy/sympy/mpmath/mptypes.pyc in
mpmathify(x, strings)
99 if hasattr(x, '_mpmath_'):
100 return mpmathify(x._mpmath_(*prec_rounding))
--> 101 raise TypeError("cannot create mpf from " + repr(x))
102
103 def try_convert_mpf_value(x, prec, rounding):
TypeError: cannot create mpf from x
thanks
Johann
On Mar 5, 4:05 pm, Ondrej Certik <[email protected]> wrote:
> On Thu, Mar 5, 2009 at 9:55 AM, johannct <[email protected]> wrote:
>
> > ok I looked at the code in solvers.py, l.606:
> > r = eq.match(a*f(x).diff(x,x) + b*diff(f(x),x) + c*f(x))
> > if r:
> > r1 = solve(r[a]*x**2 + r[b]*x + r[c], x)
> > if r1[0].is_real:
> > if len(r1) == 1:
> > return (Symbol("C1") + Symbol("C2")*x)*exp(r1[0]*x)
> > else:
> > return Symbol("C1")*exp(r1[0]*x) + Symbol("C2")*exp(r1
> > [1]*x)
> > else:
> > r2 = abs((r1[0] - r1[1])/(2*S.ImaginaryUnit))
> > return (Symbol("C2")*C.cos(r2*x) + Symbol("C1")*C.sin
> > (r2*x))*exp((r1[0] + r1[1])*x/2)
>
> > where:
> > In [19]: r
> > Out[19]:
> > ⎧ 2 ⎫
> > ⎨a: 1, b: ─, c: 1⎬
> > ⎩ x ⎭
>
> > I guess the characteristic second order test is completely irrelevant,
> > as it assumes that a,b, and c are constants, and not functions of x.
> > What is the best way to check for that?
>
> Ah, you have found a bug in Wild. These lines:
>
> a = Wild('a', exclude=[x])
> b = Wild('b', exclude=[x])
> c = Wild('c', exclude=[x])
>
> should assure, that a,b and c never contain "x". Please report it into
> a separate issue.
>
> To check, that "a" doesn't contain "x", just use:
>
> a.has(x)
>
> where x was defined on the line 593:
>
> x = f.args[0]
>
> example usage:
>
> In [1]: sin(x).has(x)
> Out[1]: True
>
> In [2]: sin(x).has(y)
> Out[2]: False
>
> Ondrej
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